Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
, here m is mass of ice and L is latent heat of fusion
So heat
So heat required to melt 1 lb of ice is equal to 151.469 KJ
Explanation:
Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :
When the tide first be at 6 ft, put h = 6 ft in above equation as :
On solving the above equation to find the value of t. It is equal to :
t = 3.551 seconds
or
t = 8.449 seconds
So, the tide of 6 ft is at 3.551 seconds and 8.449 seconds. Hence, this is the required solution.